SOLUTION: what is eight times the first of three consecutive odd integers is ten more than twice the second

Question 146850: what is eight times the first of three consecutive odd integers is ten more than twice the second
Answer by Electrified_Levi(103) (Show Source): You can put this solution on YOUR website!
Hi, Hope I can help,
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what is eight times the first of three consecutive odd integers is ten more than twice the second
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When it says consecutive odd, it means you add two to the number, then add two more to the first number(x, x+2, x+4,x+6,x+8, and so on)(you also do it to consecutive even, if it only says consecutive integers then you only add 1 every time)
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Our three consecutive odd integers are (x, x+2, x+4)
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Our equation is
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what is eight times the first of three consecutive odd integers is ten more than twice the second
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( 8x = 2(x+2) + 10 )
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We will use distribution on the right side
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( 8x = 2x + 4 + 10 )
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We will add the right side
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( 8x = 2x + 14 )
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We will move "2x to the left
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( 8x - 2x = 2x - 2x + 14 )
( 6x = 14 )
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We will divide everything by "6"
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x = , or
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We can check by replacing "x" with
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( 8x = 2(x+2) + 10 )
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First number = "x" or ( 2 )
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Second number = (x+2) or ( 4 )
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Third number = (x + 4) or (6 )
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Hope I helped, Levi
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